Which statement best reflects the solution(s) of the equation?

$\frac{x}{x-1}-\frac{1}{x-2}=\frac{2 x-5}{x^2-3 x+2}$

AThere are two solutions: $x=2$ and $x=3$.
BThere is only one solution: $x=3$. The solution $x=2$ is an extraneous solution.
CThere is only one solution: $x=3$. The solution $x=1$ is an extraneous solution.
DThere is only one solution: $x=4$. The solution $x=1$ is an extraneous solution.

Question

06-11-2026
Mathematics

Answered

Which statement best reflects the solution(s) of the equation?

$\frac{x}{x-1}-\frac{1}{x-2}=\frac{2 x-5}{x^2-3 x+2}$

AThere are two solutions: $x=2$ and $x=3$.
BThere is only one solution: $x=3$. The solution $x=2$ is an extraneous solution.
CThere is only one solution: $x=3$. The solution $x=1$ is an extraneous solution.
DThere is only one solution: $x=4$. The solution $x=1$ is an extraneous solution.

Verified Answer :

Related Questions

What was Rafael Trujillos regime really like

Choose the correct order of classifying organisms in the Linnaean system kingdomphylum class order genus family species kingdomphylumclassorderfamilygenusspecie

Usain Bolt set a world record for 100m sprint in 958 seconds Convert this time in seconds into a decimal fraction of an hour

How does using a model help me multiply decimals by whole numbers

A Spanish soldier describes Tenochtitlan The great cityhas many broad streets though among these are two or three preeminent All travel bycanoes as I have said

Poor circulation can cause cold feet True or False

A temperature of 200F is equivalent to approximately

why should you subtract 10 x 6 and not 9 x 6 or 20 x 6

Can you live without a pancreas

of the following is not true conceming the zebra mussel in the United Stateswas introduced accidentallyIt was first introduced in the Great LakesIt is still spr

GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-06T00:15:28-05:00

Factor the denominators and identify domain restrictions ($x eq 1$ and $x eq 2$).
Multiply the equation by the common denominator to simplify the expression into a quadratic equation $x^2-5x+6=0$.
Find potential roots $x=2$ and $x=3$, and verify them against the domain restrictions.
Identify $x=2$ as an extraneous solution, resulting in only one valid solution, $\boxed{x=3}$.

Explanation

Analyze the Equation and Identify Restrictions

We are given the rational equation:
\[\frac{x}{x-1} - \frac{1}{x-2} = \frac{2x-5}{x^2-3x+2}\]
First, we observe the denominators of the fractions in the equation: $x-1$, $x-2$, and the quadratic expression $x^2-3x+2$. Factoring the quadratic, we have $x^2-3x+2 = (x-1)(x-2)$.
Since division by zero is undefined, the values that make any denominator zero must be excluded from the domain. Thus, $x-1 eq 0$ and $x-2 eq 0$, which implies $x eq 1$ and $x eq 2$.

Eliminate Denominators and Simplify

To solve the equation, we eliminate the denominators by multiplying every term by the least common denominator, which is $(x-1)(x-2)$:
\[(x-1)(x-2) \left( \frac{x}{x-1} - \frac{1}{x-2} \right) = (x-1)(x-2) \left( \frac{2x-5}{(x-1)(x-2)} \right)\]
Distributing the product on the left side:
\[x(x-2) - 1(x-1) = 2x-5\]
Expanding the expressions gives:
\[x^2 - 2x - x + 1 = 2x - 5\]
Combining like terms:
\[x^2 - 3x + 1 = 2x - 5\]
Subtract $2x$ and add $5$ to both sides to form a standard quadratic equation:
\[x^2 - 5x + 6 = 0\]

Solve the Quadratic and Check for Extraneous Solutions

Now we solve the quadratic equation $x^2 - 5x + 6 = 0$. We can factor this as:
\[(x-2)(x-3) = 0\]
This yields two potential values for $x$: $x=2$ and $x=3$.
However, from our analysis in Step 1, we established the domain restrictions $x eq 1$ and $x eq 2$.
Since $x=2$ violates the domain restriction (it makes the denominator $x-2$ zero), it is an extraneous solution. The only valid solution remaining is $x=3$.

Conclusion

Comparing our findings with the given choices:

There are two solutions: $x=2$ and $x=3$. (Incorrect)
There is only one solution: $x=3$. The solution $x=2$ is an extraneous solution. (Correct)
There is only one solution: $x=3$. The solution $x=1$ is an extraneous solution. (Incorrect)
There is only one solution: $x=4$. The solution $x=1$ is an extraneous solution. (Incorrect)

Therefore, the correct statement is that there is only one solution: $x=3$, and $x=2$ is an extraneous solution. The final answer is $\boxed{\text{There is only one solution: } x=3. \text{ The solution } x=2 \text{ is an extraneous solution.}}$

Examples

This type of problem, where intermediate algebraic steps generate extraneous solutions, is crucial in fields like electrical engineering or civil engineering when modeling systems with rational functions. For example, in circuit analysis, equations involving impedances often lead to rational expressions where specific values of frequency or resistance might be mathematically obtained but physically invalid because they lead to division by zero or negative physical values. Recognizing and discarding these extraneous solutions ensures that the final model is both mathematically sound and physically applicable.